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BF
2
25 tháng 1 2020
\(n=2^{2019}-2^{2018}-...-2^1-1=2^{2019}-\left(2^{2018}+2^{2017}+...+2^1+1\right)\)
Đặt\(S=1+2+...+2^{2017}+2^{2018}\)
\(\Rightarrow2S=2+2^2+...+2^{2018}+2^{2019}\)
\(\Rightarrow2S-S=\left(2+2^2+...+2^{2018}+2^{2019}\right)-\left(1+2+...+2^{2017}+2^{2018}\right)\)
\(\Rightarrow S=2^{2019}-1\)
Mà\(n=2^{2019}-S\)
\(\Rightarrow n=2^{2019}-\left(2^{2019}-1\right)=1\)
\(\Rightarrow A=3^1+2^1+2020^1=2025\)
Happy new year :)))
XO
25 tháng 1 2020
Ta có : n = 22019 - 22018 - 22017 - .... - 22 - 2 - 1 (1)
=> 2n = 22020 - 22019 - 22018 - .... - 23 - 22 - 2 (2)
Lấy (2) trừ (1) theo vế ta có :
2n - n = (22020 - 22019 - 22018 - .... - 23 - 22 - 2) - (22019 - 22018 - 22017 - .... - 22 - 2 - 1)
=> n = 22020 - 22019 - 22019 + 1
=> n = 22020 - 2.22019 + 1 = 22020 - 22020 + 1 = 1
Khi đó A = 31 + 21 + 20201 = 3 + 2 + 2020 = 2025
Vậy A = 2025
HN
\(\left(|x|-2017\right)^{\left(n+2018\right)\cdot\left(n+2019\right)}=-\left(2^3-3^2\right)^{2019}\)
1
22 tháng 12 2018
\(\left(\left|x\right|-2017\right)^{\left(n+2018\right)\left(n+2019\right)}=-\left(2^3-3^2\right)^{2019}\)
\(\left(\left|x\right|-2017\right)^{\left(n+2018\right)\left(n+2019\right)}=-\left(-1\right)^{2019}=1\)
\(\Rightarrow\orbr{\begin{cases}\left(n+2018\right)\left(n+2019\right)=0\\\left|x\right|-2017=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\orbr{\begin{cases}n=-2018\\n=-2019\end{cases}}\\\orbr{\begin{cases}x=2018\\x=-2018\end{cases}}\end{cases}}\)
KN
5 tháng 11 2019
\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}=N\)
\(\Rightarrow M-N=0\Rightarrow\left(M-N\right)^2=0\)
MC
29 tháng 1 2019
Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)
\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)
\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)
\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)
Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)
\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.
Vậy \(\dfrac{B}{A}\) là số nguyên.
VM
19 tháng 8 2019
Câu b) bạn tham khảo tại đây nhé: Câu hỏi của Nguyễn Thái Hà - Toán lớp 6 - Học toán với OnlineMath.
Chúc bạn học tốt!
19 tháng 8 2019
a)
Có: \(\left|x-2017\right|\ge0\forall x\in Q\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-2017\right|+2018\ge2018\forall x\in Q\\\left|x-2017\right|+2019\ge2019\forall x\in Q\end{matrix}\right.\)
\(\Rightarrow\frac{\left|x-2017\right|+2018}{\left|x-2017\right|+2019}\ge\frac{2018}{2019}\forall x\in Q\\ \Rightarrow C\ge\frac{2018}{2019}\forall x\in Q\)
Vậy GTNN của C = \(\frac{2018}{2019}\)
\("="\Leftrightarrow\left|x-2017\right|=0\\ \Leftrightarrow x-2017=0\\ \Leftrightarrow x=2017\)
b) Có: \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{n^2-1}{n^2}\)
\(\Leftrightarrow S=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{n^2-1}{n^2}\\ \Leftrightarrow S=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{n^2-1}{n^2}\\ \Leftrightarrow S=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+...+1-\frac{1}{n^2}\\ \Leftrightarrow S=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Ta thấy từ 2 đến n có n-1 số hạng
\(\Rightarrow S=n-1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< n-1\left(1\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{\left(n-1\right)\cdot n}\\ \Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\\ \Rightarrow A< 1-\frac{1}{n}< 1\)
\(\Rightarrow S=n-1-A>n-1-1=n-2\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow n-2< S< n-1\)
Mà \(n\in N;n>2\)
\(\Rightarrow S\notin N\left(đpcm\right)\)
\(N=2^{2019}-2^{2018}-2^{2017}-...-2-1\)
\(=2^{2019}-\left(2^{2018}+2^{2017}+...+2+1\right)\)
Đặt \(B=1+2+...+2^{2017}+2^{2018}\)
\(\Rightarrow\) \(2B=2+2^2+...+2^{2018}+2^{2019}\)
\(\Rightarrow\) \(B=2^{2019}-1\)
\(\Rightarrow\) \(N=2^{2019}-2^{2019}+1=1\)
\(\Rightarrow\) \(A=20^1+11^1+2019^1\)
\(=20+11+2019\)
\(=2050\)
Study well ! >_<
N=\(2^{2019}-\left(1+2+.....2^{2018}\right)\)
Đặt B=\(1+2+..........+2^{2018}\)
2B=\(2+2^2+..........+2^{2019}\)
2B-B=B=\(2^{2019}-1\)
Suy ra N=\(2^{2019}-2^{2019}+1=1\)
A=20+11+2019=2050
hok tốt